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Chapter 3: Problem 24

Determine the formula weights of each of the following compounds: (a) Butyric acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH},\) which is responsible for the rotten smell of spoiled food; (b) sodium perborate, \(\mathrm{NaBO}_{3}\), a substance used as bleach; (c) calcium carbonate, \(\mathrm{CaCO}_{3},\) a substance found in marble. (c) \(\mathrm{CF}_{2} \mathrm{Cl}_{2},\) a refrigerant known as Freon; \((\mathbf{d}) \mathrm{NaHCO}_{3},\) known as baking soda and used in bread and pastry baking; \((\mathbf{e})\) iron pyrite, \(\mathrm{FeS}_{2}\) which has a golden appearance and is known as "Fool's Gold."

Short Answer

Expert verified
The formula weights for the given compounds are as follows: (a) Butyric acid, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}\): 88.12 (b) Sodium perborate, \(\mathrm{NaBO}_{3}\): 81.80 (c) Calcium carbonate, \(\mathrm{CaCO}_{3}\): 100.09 (d) Freon, \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\): 120.91 (e) Iron pyrite, \(\mathrm{FeS}_{2}\): 119.99

Step by step solution

01

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in butyric acid: - C: 12.01 - H: 1.01 - O: 16.00 Remember that there are 4 Carbon atoms, 8 Hydrogen atoms, and 2 Oxygen atoms in the formula.
02

Multiply and sum

Multiply the atomic weights by the number of atoms and then sum the results: Formula weight = (4 脳 12.01) + (8 脳 1.01) + (2 脳 16.00) = 48.04 + 8.08 + 32.00 = 88.12 (b) Sodium perborate, \(\mathrm{NaBO}_{3}\)
03

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in sodium perborate: - Na: 22.99 - B: 10.81 - O: 16.00
04

Multiply and sum

Multiply the atomic weights by the number of atoms and sum the results: Formula weight = 22.99 + 10.81 + (3 脳 16.00) = 22.99 + 10.81 + 48.00 = 81.80 (c) Calcium carbonate, \(\mathrm{CaCO}_{3}\)
05

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in calcium carbonate: - Ca: 40.08 - C: 12.01 - O: 16.00
06

Multiply and sum

Multiply the atomic weights by the number of atoms and sum the results: Formula weight = 40.08 + 12.01 + (3 脳 16.00) = 40.08 + 12.01 + 48.00 = 100.09 (d) Freon, \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\)
07

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in Freon: - C: 12.01 - F: 19.00 - Cl: 35.45
08

Multiply and sum

Multiply the atomic weights by the number of atoms and sum the results: Formula weight = 12.01 + (2 脳 19.00) + (2 脳 35.45) = 12.01 + 38.00 + 70.90 = 120.91 (e) Iron pyrite, \(\mathrm{FeS}_{2}\)
09

Identify atomic weights

Consult the periodic table to find the atomic weights of each element in iron pyrite: - Fe: 55.85 - S: 32.07
10

Multiply and sum

Multiply the atomic weights by the number of atoms and sum the results: Formula weight = 55.85 + (2 脳 32.07) = 55.85 + 64.14 = 119.99

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Atomic Weight
Atomic weight is the average mass of atoms of an element, expressed in atomic mass units (amu). To accurately determine atomic weights, scientists consider the distribution of isotopes of an element found in nature with their relative abundances. This value helps in calculating the molecular or formula weights of compounds. When we mention atomic weight, it's important to remember that it is often used interchangeably with "atomic mass," although they have slight differences. Atomic weight takes a weighted average considering all isotopes, whereas atomic mass refers directly to a singular atom's mass. For example, the atomic weight of carbon is approximately 12.01 amu, considering its isotopes like Carbon-12 and Carbon-13. This value is found directly on the periodic table, aiding in the quick calculation of molecular weights.
Formula Weights
Formula weights, also known as molecular weights, are the total weights of all atoms in a chemical formula. To compute the formula weight of a compound, first, identify all the elements present in the molecule. Then, look up each element's atomic weight on the periodic table. Multiply the atomic weight of each element by the number of times the element appears in the compound. Finally, sum all these products to get the formula weight.For instance, in sodium perborate (\(\mathrm{NaBO}_{3}\)), calculate its formula weight as follows:
  • Identify atomic weights: Na (22.99 amu), B (10.81 amu), O (16.00 amu).
  • Calculate the total weight: 22.99 + 10.81 + (3 脳 16.00) = 81.80 amu.
This systematic approach helps in determining the precise weight of various compounds, crucial for chemical reactions and stoichiometric calculations.
Periodic Table
The periodic table is a vital tool in chemistry, providing essential information about all known elements. Conceived by Dmitri Mendeleev, the periodic table arranges elements by increasing atomic number, and groups them based on similar chemical properties. Each element's square on the table displays its atomic number, chemical symbol, and often its atomic weight.By consulting the periodic table, chemists can quickly find the atomic weights necessary for calculations such as those required to determine the formula weight of a compound. For example, if calculating the formula weight of calcium carbonate (\(\mathrm{CaCO}_{3}\)), you find:
  • Calcium (Ca) at about 40.08 amu
  • Carbon (C) at about 12.01 amu
  • Oxygen (O) at about 16.00 amu
Thus, the periodic table is an indispensable resource in both educational contexts and professional chemical problem-solving.

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Most popular questions from this chapter

Consider a sample of calcium carbonate in the form of a cube measuring 2.005 in. on each edge. If the sample has a density of \(2.71 \mathrm{~g} / \mathrm{cm}^{3},\) how many oxygen atoms does it contain?

One of the most bizarre reactions in chemistry is called the Ugi reaction: \(\mathrm{R}_{1} \mathrm{C}(=\mathrm{O}) \mathrm{R}_{2}+\mathrm{R}_{3}-\mathrm{NH}_{2}+\mathrm{R}_{4} \mathrm{COOH}+\mathrm{R}_{5} \mathrm{NC} \rightarrow\) \(\mathrm{R}_{4} \mathrm{C}(=\mathrm{O}) \mathrm{N}\left(\mathrm{R}_{3}\right) \mathrm{C}\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) \mathrm{C}=\mathrm{ONHR}_{5}+\mathrm{H}_{2} \mathrm{O}\) (a) Write out the balanced chemical equation for the Ugi reaction, for the case where \(\mathrm{R}=\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}-\) (this is called the hexyl group) for all compounds. (b) What mass of the "hexyl Ugi product" would you form if \(435.0 \mathrm{mg}\) of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) was the limiting reactant?

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(3.5 \times 4.5 \times 2.5 \mathrm{~m}\). The density of air at \(26^{\circ} \mathrm{C}\) is \(0.00118 \mathrm{~g} / \mathrm{cm}^{3} .(\mathbf{b})\) If the HCN is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what mass of NaCN gives the lethal dose in the room? $$ 2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g) $$ (c) HCN forms when synthetic fibers containing Orlon \(^{\text {- }}\) or Acrilan \(^{\circledast}\) burn. Acrilan \(^{\circledast}\) has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so HCN is \(50.9 \%\) of the formula by mass. A rug measures \(3.5 \times 4.5 \mathrm{~m}\) and contains \(850 \mathrm{~g}\) of Acrilan \(^{\circledast}\) fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.

Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) where \(x\) indicates the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MgSO}_{4}\). When \(5.061 \mathrm{~g}\) of this hydrate is heated to \(250^{\circ} \mathrm{C},\) all the water of hydration is lost, leaving \(2.472 \mathrm{~g}\) of \(\mathrm{MgSO}_{4} .\) What is the value of \(x ?\)

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{C}_{7} \mathrm{H}_{16}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{Li}_{3} \mathrm{~N}(s)+\mathrm{BN}(s) \longrightarrow \mathrm{Li}_{3} \mathrm{BN}_{2}(s)\) (c) \(\mathrm{Zn}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(I)\) (d) \(\mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Ag}(s)+\mathrm{O}_{2}(g)\)
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